Find the sum, `sum_(k=1) ^(20) (1+2+3+.....+k)`

Explanation

To find the sum \( S = \sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k) \), we can follow these steps:

### Step 1: Use the formula for the sum of the first \( k \) natural numbers
The sum of the first \( k \) natural numbers is given by the formula:
\[ \sum_{n=1}^{k} n = \frac{k(k + 1)}{2} \]
Thus, we can rewrite our sum \( S \) as:
\[ S = \sum_{k=1}^{20} \frac{k(k + 1)}{2} \]

### Step 2: Factor out the constant
Since \( \frac{1}{2} \) is a constant, we can factor it out:
\[ S = \frac{1}{2} \sum_{k=1}^{20} k(k + 1) \]

### Step 3: Expand the summation
Now, we can expand \( k(k + 1) \):
\[ k(k + 1) = k^2 + k \]
Thus, we can rewrite the sum:
\[ S = \frac{1}{2} \left( \sum_{k=1}^{20} k^2 + \sum_{k=1}^{20} k \right) \]

### Step 4: Use the formulas for the sums of squares and natural numbers
We will use the following formulas:
1. The sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
2. The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

For \( n = 20 \):
- The sum of the first 20 natural numbers:
\[ \sum_{k=1}^{20} k = \frac{20(20 + 1)}{2} = \frac{20 \times 21}{2} = 210 \]

- The sum of the squares of the first 20 natural numbers:
\[ \sum_{k=1}^{20} k^2 = \frac{20(20 + 1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} \]
Calculating this:
\[ = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870 \]

### Step 5: Substitute back into the equation
Now substituting back into our equation for \( S \):
\[ S = \frac{1}{2} \left( 2870 + 210 \right) = \frac{1}{2} \times 3080 = 1540 \]

### Final Answer
Thus, the sum \( S = \sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k) = 1540 \).

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